排序算法：使用排序函數將元組中的元組與元素進行比較

Question

下面的撲克手評估器正在生成玩家持有的不同牌的排名。它在Python 2中運行良好，但它在python 3中不起作用，因爲sort函數不能再將元組與元素進行比較。我如何使它在python 3中表現得像在Python 2中一樣？

的排序需要決定哪些卡是最好的：

(1, (11, 8, 7, 6, 2)) # high card 11 (rank5) 
(1, (12, 11, 8, 7, 2)) # high card 12 (rank4) 
((3, 2), (11, 12)) # full house (triple and pair) (rank1) 
((2, 1, 1, 1), (12, 11, 7, 6)) # pair of 12 (rank3) 
((3, 1, 1), (12, 6, 3)) # three of 12 (rank2) 

如果一個整數相比一個元組，那麼元組的第一個項目應該是相關的。 Python 2中的排序功能將正確地排序上述，但在Python 3中，我收到一個錯誤消息整數不能比較元組。

如何修改鍵功能？

def keyfunction(x): 
    v= x[1] 
    return v 


def poker(hands): 
    scores = [(i, score(hand.split())) for i, hand in enumerate(hands)] 
    #print (scores) 
    winner = sorted(scores , key=keyfunction)[-1][0] 
    return hands[winner] 

def score(hand): 
    ranks = '23456789TJQKA' 
    rcounts = {ranks.find(r): ''.join(hand).count(r) for r, _ in hand}.items() 
    #print (rcounts) 
    score, ranks = zip(*sorted((cnt, rank) for rank, cnt in rcounts)[::-1]) 
    if len(score) == 5: 
     if ranks[0:2] == (12, 3): #adjust if 5 high straight 
      ranks = (3, 2, 1, 0, -1) 
     straight = ranks[0] - ranks[4] == 4 
     flush = len({suit for _, suit in hand}) == 1 
     '''no pair, straight, flush, or straight flush''' 
     score = ([1, (3,1,1,1)], [(3,1,1,2), (5,)])[flush][straight] 
    return score, ranks 

poker(['8C TS KC 9H 4S','AC TS KC 9H 4S', 'AD AS KD KS KC', '9C AD KD AC 8C', 'AC 5H 8D AD AS']) 

Answer 1

你可以簡單地變換整數表示高卡分配score成一元元組時。

... 
score = ([(1,), (3,1,1,1)], [(3,1,1,2), (5,)])[flush][straight] 
...